To solve this, we apply principles of momentum conservation and work-energy. The problem involves two stages:

1. Momentum Conservation to find the combined velocity of the bullet and wood after the collision.
2. Work-Energy Principle to determine the energy lost to friction as the combined system slides.

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Step 1: Conservation of Momentum

The momentum of the system is conserved during the collision. Before the collision:

• Momentum of bullet: $p_b = m_b v_b$,
• Momentum of wood: $p_w = 0$ (since it is at rest).

After the collision:

• Combined momentum: $p_{\text{final}} = (m_b + m_w) v_{\text{combined}}$.

Conservation of momentum: $m_b v_b = (m_b + m_w) v_{\text{combined}}$ where:

• $m_b = 0.004 \, \text{kg}$ (convert 4.00 g to kg),
• $m_w = 1.20 \, \text{kg}$.

$v_{\text{combined}} = \frac{m_b v_b}{m_b + m_w}.$

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Step 2: Work-Energy Principle

The combined system slides a distance $d = 0.350 \, \text{m}$ before coming to rest, losing all its kinetic energy to friction.

Work done by friction: $W = F_f \cdot d = \mu (m_b + m_w) g \cdot d$ where:

• $F_f = \mu (m_b + m_w) g$,
• $\mu = 0.25$ (coefficient of friction),
• $g = 9.8 \, \text{m/s}^2$ (acceleration due to gravity).

Kinetic energy of the system after the collision: $KE = \frac{1}{2} (m_b + m_w) v_{\text{combined}}^2.$

Set work equal to kinetic energy: $\mu (m_b + m_w) g \cdot d = \frac{1}{2} (m_b + m_w) v_{\text{combined}}^2.$

Simplify: $v_{\text{combined}}^2 = 2 \mu g \cdot d.$

Solve for $v_{\text{combined}}$: $v_{\text{combined}} = \sqrt{2 \mu g \cdot d}.$

Substitute the known values: $v_{\text{combined}} = \sqrt{2 (0.25) (9.8) (0.350)} = \sqrt{1.715} \approx 1.31 \, \text{m/s}.$

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Step 3: Solve for $v_b$

From momentum conservation: $v_b = \frac{(m_b + m_w)}{m_b} v_{\text{combined}}.$

Substitute values: $v_b = \frac{(0.004 + 1.20)}{0.004} (1.31) = 301.6 \, \text{m/s}.$

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Final Answer:

The bullet's speed before impact was approximately: $\boxed{301.6 \, \text{m/s}}.$

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Let me know if you'd like further clarifications or have additional questions! Here are 5 related questions you might explore:

1. What would happen to the bullet's speed if the coefficient of friction were doubled?
2. How would the result change if the bullet had a mass of 5 g instead of 4 g?
3. What other forces (besides friction) could act to stop the wood and bullet system?
4. Could the bullet emerge from the wood, and how would that affect the calculation?
5. What assumptions are we making about energy losses in this calculation?

Tip: Always double-check units when combining terms in physics problems to avoid errors!